Question

The radius of a cylinder is increasing at the rate of $2\, m / s$ and its height is decreasing at the rate of $3\, m / s$. When the radius is $3\, m$ and height is $5\, m$, then the volume of the cylinder would change at the rate of

• A. $87 \pi m^{3} / s$
• B. $33 \pi m^{3} / s$
• C. $27 \pi m^{3} / s$
• D. $15 \pi m^{3} / s$

Answer 1

Answer: (B)

Given, $\frac{d r}{d t}=2\, m / s$
$\frac{d h}{d t}=-3\, m / s$
$\therefore $ Volume of cylinder, $V=\pi r^{2} h$
$\therefore \frac{d V}{d t}=\pi\left[2 r h \frac{d r}{d t}+r^{2} \frac{d h}{d t}\right]$
At $r=3\, m$ and $h=5\, m$,
$\frac{d V}{d t}=\pi[2 \times 3 \times 5 \times 2-9 \times 3]$ [using Eq. (i)]
At $r=3\, m$ and $h=5\, m$,
$\frac{d V}{d t}=\pi[2 \times 3 \times 5 \times 2-9 \times 3]$ [using Eq. (i)]
$\Rightarrow \frac{d V}{d t}=\pi[60-27]=33 \pi$
$\Rightarrow \frac{d V}{d t}=33 \pi m^{3} / s$